Pirates, of course, are notoriously greedy – but they're also incredibly shrewd. And don't forget, they'll kill you, if given the chance.

**Sunday Puzzle #18: A Puzzle For Pirates**

This week's puzzle was proposed by reader William Murray. It's a classic puzzle, and comes in a variety of forms, but the version I've included here was widely popularized in a 1999 in Ian Stewart's "Mathematical Recreations" column for *Scientific American**. *The puzzle goes as follows:

*Ten rational pirates have gotten their hands on a hoard of 100 gold pieces and wish to divide the loot. They are democratic pirates, in their own way, and it is their custom to make such divisions in the following manner: The fiercest pirate makes a proposal about the division, and everybody votes on it, including the proposer. *

*If 50 percent or more are in favor, the proposal passes and is implemented forthwith. Otherwise the proposer is thrown overboard, and the procedure is repeated with the next fiercest pirate. *

*All the pirates enjoy throwing one of their fellows overboard, but if given a choice they prefer cold, hard cash. They dislike being thrown overboard themselves. All pirates are rational and know that the other pirates are also rational. Moreover, no two pirates are equally fierce, so there is a precise pecking order-and it is known to them all. The gold pieces are indivisible, and arrangements to share pieces are not permitted, because no pirate trusts his fellows to stick to such an arrangement. It's every man for himself. *

*What proposal should the fiercest pirate make to get the most gold? For convenience, number the pirates in order of meekness, so that the least fierce is number 1, the next least fierce number 2 and so on. The fiercest pirate thus gets the biggest number, and proposals proceed in reverse order from the top down.*

We'll be back next week with the solution – and a new puzzle! Got a great brainteaser, original or otherwise, that you'd like to see featured? E-mail me with your recommendations. (We welcome any and all submissions, but we're particularly interested in your picks for the most challenging Car Talk Puzzlers of all time – see the"Car Talk Puzzler Recommendations" section, below). As always, be sure to include "Sunday Puzzle" in the subject line!

*Art by Tara Jacoby*

## SOLUTION To Sunday Puzzle #17: The Four-Glasses Puzzle

Last week, you were asked to arrange four glasses in the same orientation on four corners of a square Lazy Susan. You were challenged to accomplish this while adhering to a specific set of rules, and to do so within a finite number of turns.

Sighted, the puzzle is dead simple – but things get complicated when one is challenged to tackle the puzzle blindfolded. Several of you were quick to identify the crux of this puzzle: Because the arrangement of the four glasses is randomized after every turn, and because you are only permitted to inspect and adjust two glasses (positioned either across from or beside one another) between randomizing events, it is possible that one of the four glasses will go untouched, indefinitely. Commenter DarkClem3 did a good job summarizing the hangup:

Any time you switch from diagonal to adjacent or vise-versa you are guaranteed to have one of your selected glasses be one you checked on your previous turn and one different, but because of the random rotation you are not guaranteed which glass that will be. You will always be left with the possibility that that last untouched glass will elude you again.

At first glance, this situation seems to preclude the possibility that the glasses can be arranged in the same orientation in a finite number of turns – i.e. without relying on chance. But it doesn't! In fact, one can guarantee that the glasses are oriented in the same direction, in a maximum of just five turns.

Several of you arrived at the solution, but I believe the first to provide a clear explanation was Elephant1232. Here is that solution. I've made some minor edits for the sake of clarity, and included a couple sketches:

I believe I have a method that accomplishes the task in a maximum of five turns, with the possibility of accomplishing it in one, two, three or four if chance is on my side.

Go get your glasses. It's much easier that way.

1) Grab either of the diagonal pairs, and orient them both to the Up position. If the bell doesn't ring, you know the other diagonal pair is either Up/Down or Down/Down.

2a) After the table spins, grab a diagonal pair. If the glasses are Up/Down or Down/Down, you know they are not the same glasses you handled in step one. Simply orient them both to the Up position. The bell will ring.

2b) If you happen to grab the pair you

originallyhandled, invert both glasses from the Up position into the Down position. If the bell doesn't ring, you can conclude that the pair you have yet to handle is oriented Up/Down. You therefore know that three of the four glasses are now oriented Down, with the last glass oriented Up:

3a) The table spins. Grab any diagonal pair. If they are Up/Down, invert the Up glass into the Down position and the bell will ring.

3b) If the pair you grab is Down/Down, invert

only one of the glassesto the Up position. There will now be two adjacent Ups and two adjacent Downs:

4a) The table spins. Grab two adjacent glasses. If they are either Up/Up or Down/Down, simply invert both glasses to the opposite position. The bell will ring.

4b) If they glasses are positioned Up/Down, reverse the orientation of both. You will now have a pair of diagonal Ups, and a pair of diagonal Downs.

5) The table spins. Grab any diagonal pair and invert the orientation of both. The bell will ring.

Notice that – in following steps 1, 2b, 3b, 4b, and 5 – one need not handle all the glasses to get all four into the same position. One only needs to handle three – which is doable, so longas one manipulate adjacent and diagonal glasses.

I will say Elephant 1232's was not the solution I was expecting. See, the I first place I encountered this puzzle was in the fourth chapter of Julian Havil's *Nonplussed! Mathematical Proof of Implausible Ideas**. *Havil borrowed the puzzle from puzzle-legend Martin Gardner, who posed the puzzle back in 1979. Gardner, in turn, was told the puzzle by a man named Robert Tappay, "who believed it to have originated in Russia."

ANYWAY, Havil claims in his book that "an initial, important observation... is that the selection of the [glasses] has essentially two forms: a side pair or a diagonal pair." No complaints there. He then, however, claims that "it is also clear that these choices must alternate, otherwise we could go on repeating ourselves forever." The solution he provides is the same one I arrived at, the first time I tried my hand at this puzzle:

- Choose any diagonal pair and orient them to be the same way up.
- If the bell does not ring, spin the table and choose two adjacent glasses. If both are burned up, leave them. Otherwise, invert the glass that is turned down. If the bell doesn't ring, you know (as we did above) that three of the glasses are in the same orientation.
- Spin table and choose a diagonal pair of glasses. If one is turned down, invert it. The bell will ring. If both are turned up, invert one of them. As above, the resulting configuration is two adjacent Ups and two adjacent Downs.
- Spin the table. Select two adjacent glasses and invert them. If they were both oriented in the same position, the bell will ring. Otherwise, you will have a diagonal pair in the Up position and a diagonal pair in the Down position.
- Spin the table. Select a diagonal pair of glasses and invert them. The bell will ring.

Notice how, in Havil's solution, the choice between adjacent and diagonal pairs alternates? As far as I can tell, this is not actually necessary, as Elephant1232's solution demonstrates! (But feel free to e-mail me if this doesn't check out on your end.)

Havil goes on to explore some generalizations of the problem involving increasingly numerous glasses and increasingly numerous hands with which to manipulate them. It's too lengthy to include here, but if you're mathematically inclined, I encourage you to grab a copy for yourself.

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## DISCUSSION

Ok more serious answer.

the proposed scenario is that the votes would look something like this

In this scenario everyone is better off than they would have been in the next vote if they threw a crew member overboard.

But there's a problem with this, what if they're all greedy bastards?

If there was a negotiation period then whoever controls the vote could demand a larger share of the booty. In this scenario at vote 8, pirate 10 could demand the whole lot, or he (or she) will throw pirate 8 overboard. With this in mind the scenario might look like this.

But of course there is no negotiation period, but all the pirates, being rational, would be aware of what would happen if

Now pirate one has to choose one of the two scenarios. Being a greedy bastard he may choose scenario 1 and keep most of the gold, giving away only four coins. Now the other 8 pirates are also all aware that there are two potential scenarios. For 8 of those pirates it doesn't make a difference, either scenario in round 2 they get nothing. Apart from pirate 9. If pirate 1 offers him 1 gold coin, and he says nay instead of yay, he throws the whole vote and pirate 1 gets thrown to the sharks. Now pirate 2 knows that 1 gold coin is not enough to keep everyone happy and to keep himself alive. So he now HAS to give away all his gold (25 pieces each) to pirates 4, 6, 9 and 10, because we're now in scenario 2 and even though there's no explicit negotiation there's an implicit negotiation.

But starting again pirate 1 knows therefore that offering pirate 9 1 coin is not going to be enough, because there are better options for him if he votes nay, so his proposal will from the start try to buy out pirate 9, the one pirate that can screw up his plans without sacrificing any gold.

But now with this knowledge pirate 9 controls the whole vote so pirate 1 needs to offer him more than he would get by voting nay.

so the actual proposal by pirate 1 would be and would be accepted.